Question

A canoe has a velocity of 0.310 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.460 m/s east relative to the earth.

a. Find the magnitude of the velocity Vc/r of the canoe relative to the river.

b. Find the direction of the velocity of the canoe relative to the river. Express your answer as an angle measured south of west.

v⃗ c/r

Answer 1

Given values are:

Vc/e = Velocity of canoe w.r.t. earth = 0.310 m/sec in south east direction

Vc/e = 0.310*cos 45 deg i - 0.310*sin 45 deg j

Vc/e = 0.219 i - 0.219 j

Vr/e = Velocity of river w.r.t. earth = 0.460 m/sec in east direction

Vr/e = 0.460 i

Now we need to find

Vc/r = Velocity of canoe w.r.t. river

So Using relative velocity concept

Vc/r = Vc/e - Vr/e

Vc/r = (0.219 i - 0.219 j) - (0.460 i)

Vc/r = -0.241 i - 0.219 j

Magnitude of Vc/r will be:

|Vc/r| = sqrt ((-0.241)^2 + (-0.219)^2)

|Vc/r| = 0.326 m/sec

Direction of Vc/r will be:

Since Vc/r is in 3rd quadrant so direction will be south of west and angle will be:

Angle = arctan ((Vc/r)_y/(Vc/r)_x)

Angle = arctan (0.219/0.241) = 42.26 deg

Direction of Vc/r = 42.3 deg South of west

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